Statistical Mechanics I Fall 2005 Mid-Term Quiz Review Problems

8.333: Statistical Mechanics I Fall 2005 Mid-term Quiz� Review Problems The Mid-term quiz will take place on Monday 10/24/05 in room 1-190 from 2:30 to 4:00 pm. There will be a recitation with quiz review on Friday 10/21/05. All topics up to (but not including) the micro-canonical ensemble will be covered. The exam is ‘closed book,’ but if you wish you may bring a two-sided sheet of formulas. The enclosed exams (and solutions) from the previous years are intended to help you review the material. Solutions to the midterm are available in the exam s section . ******** Answer all three problems, but note that the ﬁrst parts of each problem are easier than its last parts. Therefore, make sure to proceed to the next problem when you get stuck. You may ﬁnd the following information helpful:

Physical Constants 31 27 Electron mass me 9.1 10− Kg Proton mass mp 1.7 10− Kg ∝ × 19 ∝ × 34 1 Electron Charge e 1.6 10− C Planck’s const./2α h¯ 1.1 10− J s− ∝ × 8 1 ∝ × 8 2 4 Speed of light c 3.0 10 ms− Stefan’s const. δ 5.7 10− W m− K− ∝ × 23 1 ∝ × 23 1 Boltzmann’s const. k 1.4 10− J K− Avogadro’s number N 6.0 10 mol− B ∝ × 0 ∝ × Conversion Factors 5 2 10 4 1atm 1.0 10 N m− 1A˚ 10− m 1eV 1.1 10 K ∞ × ∞ ∞ ×

Thermodynamics dE = T dS +dW¯ For a gas: dW¯ = P dV For a wire: dW¯ = J dx −

Mathematical Formulas �λ � n �x n! 1 0 dx x e− = �n+1 2! = 2

⎪ 2 2 2 � � � x ⇒ 2 β k dx exp ikx 2β2 = 2αδ exp 2 limN ln N ! = N ln N N −� − − − �� − � � � � ⎪ n n ikx ( ik) n ikx ( ik) n e− = � − x ln e− = � − x n=0 n! � ≡ n=1 n! � ≡c � � � 2 4 � � � 3 5 cosh(x) = 1 + x + x + sinh(x) = x + x + x + 2! 4! ··· 3! 5! ···

2λd/2 Surface area of a unit sphere in d dimensions Sd = (d/2 1)! −

1 8.333: Statistical Mechanics I� Fall 1999 Mid-term Quiz�

1. Photon gas Carnot cycle: The aim of this problem is to obtain the blackbody ra diation relation, E(T, V ) V T 4, starting from the equation of state, by performing an → inﬁnitesimal Carnot cycle on the photon gas. P

P+dP T+dT P T V V+dV V (a) Express the work done, W , in the above cycle, in terms of dV and dP . Ignoring higher order terms, net work is the area of the cycle, given by W = dP dV . • (b) Express the heat absorbed, Q, in expanding the gas along an isotherm, in terms of P , dV , and an appropriate derivative of E(T, V ). Applying the ﬁrst law, the heat absorbed is •

γE� γE γE Q = dE + P dV = dT + dV + P dV = + P dV. γT γV γV �� V� � �T �isotherm �� T �

(c) Using the eﬃciency of the Carnot cycle, relate the above expressions for W and Q to T and dT . The eﬃciency of the Carnot cycle (σ = dT/T ) is here calculated as • W dP dT σ = =� = . Q [(γE/γV )T + P ] T

(d) Observations indicate that the pressure of the photon gas is given�by P = AT 4 , 2 4 3 where A = α kB /45 (¯hc) is a constant. Use this information to obtain E(T, V ), assuming E(0, V ) = 0. From the result of part (c) and the relation P = AT 4 , • � γE γE 4AT 4 = + AT 4 , or = 3AT 4 , γV� γV � �T� � �T so that E = 3AV T 4 .

2 (e) Find the relation describing the adiabatic paths in the above cycle. Adiabatic curves are given by dQ = 0, or •

γE γE 0 = dT + dV + P dV = 3V dP + 4P dV, γT γV � �V � �T i.e. P V 4/3 = constant. ******** 2. Moments of momentum: Consider a gas of N classical particles of mass m in thermal equilibrium at a temperature T , in a box of volume V .

(a) Write down the equilibrium one particle density feq. (p,τ τ q ), for coordinate τq, and mo mentum pτ. The equilibrium Maxwell-Boltzmann distribution reads •

n p2 f (p,τ τ q ) = exp . 3/2 − 2mk T (2αmkB T ) � B �

(b) Calculate the joint characteristic function, exp iτk pτ , for momentum. − · Performing the Gaussian average yields � � �� •

iγk pγ mkB T 2 p˜ τk = e− · = exp k . − 2 � � � � � �

(c) Find all the joint cumulants pσ pmp n . x y z c The cumulants are calculated from the characteristic function, as � � • σ m n γ γ γ pσpm p n = ln p˜ τk x y z c γ ( ikx) γ ( iky ) γ ( ikz ) � �− � �− � � − � � ��γk=0 � � � = mkB T (βσ2βm0βn0 + βσ0βm2βn0 + βσ0βm0βn2) , � � i.e., there are only second cumulants; all other cumulants are zero. (d) Calculate the joint moment p p (pτ p τ ) . � � � · ≡ Using Wick’s theorem • p p (pτ p τ ) = p p p p � � � · ≡ � � � π π ≡ = p p p p + 2 p p p p � � � ≡ � π π ≡ � � π ≡ � � π ≡ 2 2 = (mkB T ) β�� βππ + 2 (mkB T ) β�π β�π 2 = 5 (mkB T ) β�� .

3 Alternatively, directly from the characteristic function,

γ γ γ γ p�p� (pτ pτ ) = p˜ τk � · ≡ γ ( ik ) γ ( ik ) γ ( ik ) γ ( ik ) � � π π �γk=0 − − − − � � � γ γ 2 mkB T γ 2 τ 2 � 2 k = 3mkB (mkB T ) k e�− γ ( ik ) γ ( ik ) − − � − � �γk=0 2 � � � = 5 (mkBT ) β��. � � ******** 3. Light and matter: In this problem we use kinetic theory to explore the equilibrium between atoms and radiation.

(a) The atoms are assumed to be either in their ground state a0, or in an excited state a1, which has a higher energy π. By considering the atoms as a collection of N ﬁxed two-state systems of energy E (i.e. ignoring their coordinates and momenta), calculate the ratio

n1/n0 of densities of atoms in the two states as a function of temperature T . The energy and temperature of a two-state system are related by • N η E = , 1 + exp (η/kB T ) leading to

N E/η N exp (η/kB T ) E/η N 1 n0 =− = , and n1 = = , VV 1 + exp (η/kBT ) VV 1 + exp (η/kB T )

so that n1 η = exp . n −k T 0 �B � Consider photons θ of frequency � = π/h¯ and momentum pτ =h�/c ¯ , which can | | interact with the atoms through the following processes: (i) Spontaneous emission: a a + θ. 1 � 0 (ii) Adsorption: a + θ a . 0 � 1 (iii) Stimulated emission: a + θ a + θ + θ. 1 � 0 Assume that spontaneous emission occurs with a probability δsp, and that adsorption and stimulated emission have constant (angle-independent) diﬀerential cross-sections of

δad/4α and δst/4α, respectively. (b) Write down the Boltzmann equation governing the density f of the photon gas, treating

the atoms as ﬁxed scatterers of densities n0 and n1. The Boltzmann equation for photons in the presence of ﬁxed scatterers reads • γf γf + pτ = δadn0cf + δstn1cf + δspn1. γt · γτq −

4 (c) Find the equilibrium density feq. for the photons of the above frequency. In uniform equilibrium, the left-hand side vanishes, leaving • δ n cf + δ n cf + δ n = 0, − ad 0 eq. st 1 eq. sp 1 i.e. 1 δsp 1 δsp feq. = = . c δ n /n δ c δ exp (η/k T ) δ ad 0 1 − st ad B − st

(d) According to Planck’s law, the density of photons at a temperature T depends on their 1 3 frequency � as f = [exp (¯h�/k T ) 1]− /h . What does this imply about the above eq. B − cross sections? The result of part (c) agrees with Planck’s law if • c δ = δ , and δ = δ , ad st sp h3 st a conclusion first reached by Einstein, and verified later with explicit quantum mechanical calculations of cross-sections. (e) Consider a situation in which light shines along the x axis on a collection of atoms whose boundary coincides with the x = 0 plane, as illustrated in the figure.

�

vacuum matter (n0 , n1 )

Clearly, f will depend on x (and px), but will be independent of y and z. Adapt the Boltzmann equation you propose in part (b) to the case of a uniform incoming ﬂux of photons with momentum pτ =h� ¯ x/cˆ . What is the penetration length across which the incoming ﬂux decays? In this situation, the Boltzmann equation reduces to •

γf n1 px = δstc (n1 n0) f + χ (x) . γx − h3 � � To the uniform solution obtained before, one can add an exponentially decaying term for x > 0, i.e. ax/px f (px, x > 0) = A (px) e− + feq. (px) .

5 The constant A (px) can be determined by matching to solution for x < 0 at x = 0, and is related to the incoming ﬂux. The penetration depth d is the inverse of the decay parameter, and given by px d = , with a = δstc (n0 n1) > 0. a − ********

6� 8.333: Statistical Mechanics I Fall 2000 Mid-term Quiz�

1. Superconducting transition: Many metals become superconductors at low temperatures T , and magnetic ﬁelds B. The heat capacities of the two phases at zero magnetic ﬁeld are approximately given by

3 Cs(T ) = V �T in the superconducting phase , 3 � Cn (T ) = V ωT + θT in the normal phase where V is the volume, and� �, ω, θ ⎬are constants. (There is no appreciable change in { } volume at this transition, and mechanical work can be ignored throughout this problem.)

(a) Calculate the entropies Ss(T ) and Sn(T ) of the two phases at zero field, using the third law of thermodynamics. Finite temperature entropies are obtained by integrating dS = dQ/T¯ , starting from • S(T = 0) = 0. Using the heat capacities to obtain the heat inputs, we find � 3 � 3 dSs �T Cs = V �T = T , = Ss = V , dT ≥ 3 . � 3 ⎧ 3 dSn ωT ⎧ Cn = V ωT + θT = T , = Sn = V + θT � dT ≥ 3 � � ⎧ � ⎬ ⎭⎧ (b) Experiments indicate that there is no latent heat (L = 0) for the transition between the normal and superconducting phases at zero field. Use this information to obtain the

transition temperature Tc, as a function of �, ω, and θ. The Latent hear for the transition is related to the diﬀerence in entropies, and thus • L = T (S (T ) S (T )) = 0. c n c − s c Using the entropies calculated in the previous part, we obtain 3 3 �T cωT c 3θ = + θTc, = Tc = . 3 3 ≥ � ω � −

(c) At zero temperature, the electrons in the superconductor form bound Cooper pairs. As a result, the internal energy of the superconductor is reduced by an amount V �, i.e. E (T = 0) = E and E (T = 0) = E V � for the metal and superconductor, respectively. n 0 s 0 − Calculate the internal energies of both phases at ﬁnite temperatures. Since dE = T dS + BdM + µdN , for dN = 0, and B = 0, we have dE = T dS = CdT . • Integrating the given expressions for heat capacity, and starting with the internal energies

E0 and E0 V � at T = 0, yields − � 4 Es(T ) = E0 + V � + T − 4 � � . � ω 4 θ 2 ⎧ En(T ) = E0 + V T + T � 4 2 � � ⎧ ⎭ 7 (d) By comparing the Gibbs free energies (or chemical potentials) in the two phases, obtain an expression for the energy gap � in terms of �, ω, and θ. The Gibbs free energy G = E T S BM = µN can be calculated for B = 0 in each • − − phase, using the results obtained before, as

� 4 � 3 � 4 Gs(T ) = E0 + V � + T T V T = E0 V � + T − 4− 3 − 12 � � � � . � ω 4 θ 2 ω 3 ω 4 θ 2 ⎧ Gn(T ) = E0 + V T + T T V T + θT = E0 V T + T � 4 2 − 3 − 12 2 � � � � � � ⎭⎧ At the transition point, the chemical potentials (and hence the Gibbs free energies) must be equal, leading to

� ω θ θ � ω � + T 4 = T 4 + T 2 , = � = T 2 − T 4 . 12 c 12 c 2 c ≥ 2c − 12 c

Using the value of T = 3θ/(� ω), we obtain c − � 3 θ2 � = . 4 � ω −

(e) In the presence of a magnetic ﬁeld B, inclusion of magnetic work results in dE = T dS +BdM +µdN, where M is the magnetization. The superconducting phase is a perfect diamagnet, expelling the magnetic ﬁeld from its interior, such that M = V B/(4α) in s − appropriate units. The normal metal can be regarded as approximately non-magnetic,

with Mn = 0. Use this information, in conjunction with previous results, to show that the superconducting phase becomes normal for magnetic ﬁelds larger than

T 2 Bc(T ) = B0 1 , − T 2 � c �

giving an expression for B0. Since dG = SdT MdB + µdN, we have to add the integral of MdB to the Gibbs • − − − free energies calculated in the previous section for B = 0. There is no change in the metallic phase since M = 0, while in the superconducting phase there is an additional n contribution of M dB = (V/4α) BdB = (V/8α)B2 . Hence the Gibbs free energies − s at finite field are ⎪ ⎪ 2 � 4 B Gs(T, B) = E0 V � + T + V − 12 8α � � � . ⎧ ω 4 θ 2 ⎧ Gn(T, B) = E0 V T + T � − 12 2 � � ⎧ ⎭ 8� Equating the Gibbs free energies gives a critical magnetic field

B2 θ � ω 3 θ2 θ � ω c = � T 2 + − T 4 = T 2 + − T 4 8α − 2 12 4 � ω − 2 12 − 2 2 � ω 3θ 6θT � ω 2 = − + T 4 = − T 2 T 2 , 12 � ω − � ω 12 c − ��− � − � � �

where we have used the values of � and Tc obtained before. Taking the square root of the above expression gives

2 2 T 2α(� ω) 2 6αθ Bc = B0 1 , where B0 = − T = = Tc 2αθ. − T 2 3 c � ω � c � � � − �

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2. Probabilities: Particles of type A or B are chosen independently with probabilities pA and pB .

(a) What is the probability p(NA,N) that NA out of the N particles are of type A? The answer is the binomial probability distribution •

N! NA N NB p(NA,N) = p p − . N !(N N )! A B A − A

(b) Calculate the mean and the variance of NA. We can write • N nA = ti, i=1 � where ti = 1 if the i-th particle is A, and 0 if it is B. The mean value is then equal to

N N N = t = (p 1 + p 0) = Np . � A≡ � i ≡ A × B × A i=1 i=1 � � Similarly, since the t are independent variables, { i} N N N 2 = t2 t 2 = p p2 = Np p . A c i − � i≡ A − A A B i=1 i=1 � � � ��

9 According to the central limit theorem the PDF of the sum of independent variables • for large N approaches a Gaussian of the right mean and variance. Using the mean and variance calculated in the previous part, we get

2 (NA NpA) 1 lim p(NA,N) exp − . N 1 2Np p ⇒ ∞ ∝ �− A B � 2αNpApB

(d) Apply Stirling’s approximation (ln N! N ln N N) to ln p(N ,N) [using the prob ∝ − A ability calculated in part (a), not part (c)] to ﬁnd the most likely value, N , for N 1. A √ Applying Stirling’s approximation to the logarithm of the binomial distribution gives •

ln p(NA,N) = ln N! ln NA! ln(N NA)! + NA ln pA + (N NA) ln pB − − − − NA NA NA ln (N NA) ln 1 + NA ln pA + (N NA) ln pB. ∝ − N − − − N − � � � �

The most likely value, NA, is obtained by setting the derivative of the above expression

with respect to NA to zero, i.e.

d ln p NA N pA = ln + ln = 0, = NA = pAN. dNA − N N N pB ≥ � − A � Thus the most likely value is the same as the mean in this limit. (e) Expand ln p(N ,N) calculated in (d) around its maximum to second order in A N N , and check for consistency with the result from the central limit theorem. A − A Taking a second derivative of ln p gives •� � d2 ln p 1 1 N 1 2 = = = . dN −N − N N −N N N −NpApB A A − A A − A � � The expansion of ln p around its maximum thus gives

2 (NA pAN) ln p − , ∝ − 2NpApB

which is consistent with the result from the central limit theorem. The correct normaliza tion is also obtained if the next term in the Stirling approximation is included. ********

3. Thermal Conductivity: Consider a classical gas between two plates separated by a

distance w. One plate at y = 0 is maintained at a temperature T1, while the other plate at

10� y = w is at a diﬀerent temperature T2. The gas velocity is zero, so that the initial zeroth order approximation to the one particle density is,

n(y) pτ pτ f 0(p,τ x, y, z) = exp · . 1 3/2 −2mk T (y) [2αmkB T (y)] � B �

(a) What is the necessary relation between n(y) and T (y), to ensure that the gas velocity τu remains zero? (Use this relation between n(y) and T (y) in the remainder of this problem.) Since there is no external force acting on the gas between plates, the gas can only ﬂow • locally if there are variations in pressure. Since the local pressure is P (y) = n(y)kB T (y), the condition for the ﬂuid to be stationary is

n(y)T (y) = constant.

(b) Using Wick’s theorem, or otherwise, show that

0 0 p 2 p p 0 = 3 (mk T ) , and p 4 p p p p 0 = 15 (mk T )2 , ∞ � � �≡ B ∞ � � � � � ≡ B 0 where � �0 indicates local averages with the Gaussian� � weight f 0 . Use the result p6 = �O≡ 1 105(mk T )3 (you don’t have to derive this) in conjunction with symmetry arguments to B � � conclude 2 4 0 3 py p = 35 (mkB T ) . � � The Gaussian weight has a covariance p p 0 = β (mk T ). Using Wick’s theorem • � � � ≡ �� B gives 0 p 2 = p p 0 = (mk T ) β = 3 (mk T ) . � � �≡ B �� B Similarly � �

0 p 4 = p p p p 0 = (mk T )(2 β + 2β β ) = 15 (mk T )2 . � � � � � ≡ B �� B The symmetry� �along the three directions implies

2 4 0 2 4 0 2 4 0 1 2 4 0 1 3 3 p p = p p = p p = p p = 105 (mkBT ) = 35 (mkBT ) . x y z 3 3 × � � � � � � � � (c) The zeroth order approximation does not lead to relaxation of temperature/density 1 variations related as in part (a). Find a better (time independent) approximation f1 (p,τ y), by linearizing the Boltzmann equation in the single collision time approximation, to

1 0 γ py γ f f f 1 + f 0 1 − 1 , L 1 ∝ γt m γy 1 ∝ − ε � � K � ⎬ 11 where εK is of the order of the mean time between collisions. Since there are only variations in y, we have • 2 γ py γ 0 0 py 0 0 py 3 p 3 + f = f γy ln f = f γy ln n ln T ln (2αmkB) γt m γy 1 1 m 1 1 m − 2− 2mk T − 2 � � � B � 2 2 py γyn 3 γyT p γT py 5 p γyT = f 0 + = f 0 + , 1 m n − 2 T 2mk TT 1 m −2 2mk TT � B � � B � where in the last equality we have used nT = constant to get γ n/n = γ T /T . Hence y − y the ﬁrst order result is 2 1 0 py p 5 γy T f (p,τ y) = f ( p,τ y) 1 εK . 1 1 − m 2mk T − 2 T � � B � �

1 (d) Use f1 , along with the averages obtained in part (b), to calculate hy , the y component of the heat transfer vector, and hence find K, the coefficient of thermal conductivity. Since the velocity τu is zero, the heat transfer vector is • 2 1 mc n 2 1 hy = n cy = pyp . 2 2m2 � � In the zeroth order Gaussian weight all odd moments �of p have� zero average. The correc 1 tions in f1 , however, give a non-zero heat transfer 0 n γ T p p2 5 h = ε y y p p2 . y K 2 y − 2m T m 2mkB T − 2 � � � � 2 4 0 2 2 0 Note that we need the Gaussian averages of py p and py p . From the results of part 3 2 (b), these averages are equal to 35(mkB T ) �and 5(�mkB T� ) , respectively.� Hence 2 n γy T 2 35 5 5 5 nεK kB T hy = εK (mkBT ) × = γyT. − 2m3 T 2 − 2 −2 m � � The coefficient of thermal conductivity relates the heat transferred to the temperature gradient by τh = K T , and hence we can identify − ≈ 2 5 nεKk T K = B . 2 m

(e) What is the temperature proﬁle, T (y), of the gas in steady state? Since γ T is proportional to γ h , there will be no time variation if h is a constant. • t − y y y But h = Kγ T , where K, which is proportional to the product nT , is a constant in y − y the situation under investigation. Hence γy T must be constant, and T (y) varies linearly between the two plates. Subject to the boundary conditions of T (0) = T1, and T (w) = T2, this gives T2 T1 T (y) = T1 + − y. w

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12� 8.333: Statistical Mechanics I Fall 2003 Mid-term Quiz�

1. Hard core gas: A gas obeys the equation of state P (V Nb) = Nk T , and has a − B heat capacity CV independent of temperature. (N is kept ﬁxed in the following.) (a) Find the Maxwell relation involving γS/γV . |T ,N For dN = 0, •

γS γP d(E TS) = SdT P dV, = = . − − − ≥ γV γT �T ,N �V,N � � � � � � (b) By calculating dE(T, V ), show that E is a function of T (and N) only. Writing dS in terms of dT and dV , •

γS γS dE = T dS P dV = T dT + dV P dV. − γT γV − � �V,N �T ,N � � � � � Using the Maxwell relation from part (a), we� ﬁnd �

γS γP dE(T, V ) = T dT + T P dV. γT γT − �V,N � �V,N � � � � � But from the equation of state, we get� �

NkBT γP P γS P = , = = , = dE(T, V ) = T dT, (V Nb) ≥ γT T ≥ γT �V,N �V,N − � � � � i.e. E(T, N, V ) = E(T, N) does not depend� on V . � (c) Show that θ C /C = 1 + Nk /C (independent of T and V ). ∞ P V B V The heat capacity is •

γQ γE + P γV γE γV CP = = = + P. γT γT γT γT �P �P �P �P � � � � � � � � But, since E = E(T ) only, � � � � γE γE = = CV , γT γT �P �V � � and from the equation of state we get� � � �

γV NkB NkB = , = CP = CV + NkB, = θ = 1 + , γT P ≥ ≥ C �P V � � � 13 which is independent of T , since CV is independent of temperature. The independence of CV from V also follows from part (a). (d) By writing an expression for E(P, V ), or otherwise, show that an adiabatic change satisﬁes the equation P (V N b)π =constant. − Using the equation of state, we have •

P (V N b) CV dE = CV dT = CV d − = (P dV + (V N b)dP ) . N k N k − � B � B The adiabatic condition, dQ = dE + P dV = 0, can now be written as

CV CV 0 = dQ = 1 + P d(V N b) + (V N b)dP. N k − N k − � B � B Dividing by C P (V N b)/(N k ) yields V − B dP d(V N b) + θ − = 0, = ln [P (V N b)π ] = constant. P (V N b) ≥ − −

******** 2. Energy of a gas: The probability density to ﬁnd a particle of momentum p ∞ (px, py , pz ) in a gas at temperature T is given by

1 p2 p(p) = exp , where p 2 = p p . 3/2 − 2mk T · (2αmkB T ) � B �

(a) Using Wick’s theorem, or otherwise, calculate the averages p2 and (p p)(p p) . � · · ≡ From the Gaussian form we obtain p p = mk T β , where � and ω label any of the • � � � ≡ B �� three components of the momentum. Therefore:

p 2 = p p = mk T β = 3mk T, � � �≡ B �� B and using Wick’s theorem� �

(p p)(p p) = p p p p = (mk T )[2 β β + 2β β ] = 15 (mk T )2 . � · · ≡ � � � � � ≡ B �� B

(b) Calculate the characteristic function for the energy π = p2/2m of a gas particle.

The characteristic function π is the average eikα , which is easily calculated by Gaussian • integration as � � 3 2 ikα ikp 2 /2m d p 1 p 3/2 e = e = exp ik = (1 ikkBT )− . 3/2 − k T 2m − � (2αmkB T ) �� B � � � � � � 14� (c) Using the characteristic function, or otherwise, calculate the mth cumulant of the particle energy πm . � ≡c The cumulants are obtained from the expansion • m � m � ikα (ik) m 3 3 (kB T ) m ln e = π = ln (1 ikkBT ) = (ik) , m! � ≡c −2 − 2 m m=1 m=1 � � � � as m 3 m π = (m 1)! (kBT ) . � ≡c 2 −

N (d) The total energy of a gas of N (independent) particles is given by E = i=1 πi, where π is the kinetic energy of the ith particle, as given above. Use the central limit theorem i � to compute the probability density for energy, p(E), for N 1. √ Since the energy E is the sum of N identically distributed independent variables, its • cumulants are simply N times those for a signle variable, i.e.

m m 3 m E = N π = N(m 1)! (kBT ) . � ≡c � ≡c 2 − According to the central limit theorem, in the large N limit the mean and variance are suﬃcient to describe the probability density, which thus assumes the Gaussian form

2 1 (E 3NkBT/2) p(E) = exp − . ⇒3αNkB T �− 3NkB T �

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3. ‘Relativistic’ gas: Consider a gas of particles with a ‘relativistic’ one particle Hamil tonian = c p , where p = p2 + p2 + p2 is the magnitude of the momentum. (The H1 | | | | x y z external potential is assumed to� be zero, expect at the edges of the box conﬁning the gas particles.) Throughout this problem treat the two body interactions and collisions precisely as in the case of classical particles considered in lectures.

(a) Write down the Boltzmann equation for the one-particle density f1(p, q, t), using the same collision form as employed in lectures (without derivation). The Boltzmann equation has the general form • f = C[f , f ]. L 1 1 1 The collision term is assumed to be the same as in the classical case derived in lectures, and thus given by

3 2 C[f , f ] = d p2d b v2 v1 [f (p1)f (p2) f (p→1)f (p→2)] . 1 1 − | − | 1 1 − 1 1 � 15 (There are various subtleties in treatment of relativistic collisions, such as the meaning of v2 v1 , which shall be ignored here.) The streaming terms have the form | − |

γ 1 p� f1(p, q, t) = γtf1 + 1, f1 = γt + H γ� f1 = γt + c γ� f1. L {H } γp p � � � � | | �

(b) The two body collisions conserve the number of particles, the momentum, and the particle energies as given by . Write down the most general form f 0(p, q, t) that sets H1 1 the collision integrand in the Boltzmann equation to zero. (You do not need to normalize this solution.) The integrand in C[f , f ] is zero if at each q, ln f (p1)+ln f (p2) = ln f (p→1)+ln f (p→2). • 1 1 1 1 1 1 This can be achieved if ln f1 = µ aµ (q, t)λµ(p), where λµ(p) are quantities conserved in a two body collision, and a are functions independent of p. In our case, the conserved µ � quantities are 1 (particle number), p (momentum), and c p (energy), leading to | | 0 f (p, q, t) = exp [ a (q, t) a1(q, t) p a (q, t)c p ] . 1 − 0 − · − 2 | |

For any function λ(p) which is conserved in the collisions, there is a hydrodynamic equation of the form

p� p� γt (n λ ) + γ� n c λ n γtλ n c γ�λ = 0, � ≡ p − � ≡ − p � �| | �� | | � 3 where n(q, t) = d pf1(p, q, t) is the local density, and

⎪ 1 = d3 pf (p, q, t) . �O≡ n 1 O � (c) Obtain the equation governing the density n(q, t), in terms of the average local velocity u = cp / p . � � � | |≡ Substituting λ = 1 in the conservation equation gives • γ n + γ (nu ) = 0, with u = cp / p . t � � � � � | |≡

(d) Find the hydrodynamic equation for the local momentum density α (q, t) p , in � ∞ � �≡ terms of the pressure tensor P = nc (p α ) (p α ) / p . �� − � � − � | |≡ Since momentum is conserved in the collisions, we can obtain a hydrodynamic equation • by putting λ = p α in the general conservation form. Since λ = 0, this leads to � � − � � �≡

λ� + α� γ� n c λ� + nγtα� + nu�γ�α� = 0. p � � | | �� 16� Further simpliﬁcation and rearrangements leads to�

1 1 λ� Dtα� γtα� + u�γ�α� = γ�P�� γ� nα�c . ∞ −n − n p � �| |�� (Unfortunately, as currently formulated, the problem does not lead to a clean answer, in that there is a second term in the above result that does not depend on P�� .)

(e) Find the (normalized) one particle density f1(p, q, t) for a gas of N such particles in a box of volume V , in equilibrium at a temperature T . At equilibrium, the temperature T and the density n = N/V are uniform across the • system, and there is no local velocity. The general form obtained in part (b) now gives

3 N c p 1 c f 0(p, q, t) = exp | | . 1 V −k T 8α k T � B � �B � 3 3 The normalization factor is obtained by requiring N = V d pf1, noting that d p = 2 n p/a n+1 4αp dp, and using � dpp e− = n!a . 0 ⎪ (f) Evaluate the pressure⎪ tensor P�� for the above gas in equilibrium at temperature T . For the gas at equilibrium α = u = 0, and the pressure tensor is given by • � �

p�p� px px nc p p P�� = nc = ncβ�� = β�� · . p p 3 p � | | � � | | � � | | � In rewriting the above equation we have taken advantage of the rotational symmetry of the system. The expectation value is simply

� 2 cp/kB T 0 dpp pe− kB T p = = 3 , �| |≡ � dpp2e cp/kB T c ⎪ 0 − leading to ⎪

P�� = β�� nkB T, which is the usual formula for an ideal gas. ********

17� 8.333: Statistical Mechanics I Fall 2004 Mid-term Quiz�

1. Wire: Experiments on stretching an elastic wire indicate that, at a temperature T , a displacement x requires a force

J = ax bT + cT x, − where a, b, and c are constants. Furthermore, its heat capacity at constant displacement

is proportional to temperature, i.e. Cx = A(x)T . (a) Use an appropriate Maxwell relation to calculate γS/γx . |T From dF = SdT + Jdx, we obtain • −

γS γJ = = b cx. γx − γT − �T �x � � � � � � (b) Show that A has to be independent of x, i.e. dA/dx = 0.

We have C = T εS = A(x)T , where S = S(T, x). Thus • x εT x � � γA γ γS γ γS = = = 0 γx γx γT γT γx from part (a), implying that A is independent of x. (c) Give the expression for S(T, x), and comment on whether it is compatible with the third law of thermodynamics. By integrating the derivatives of S given above, S(x, T ) can be calculated as • � � T =T x =x γS(T →, x = 0) γS(T, x→) S(x, T ) = S(0, 0) + dT → + dx

� γT � γx �T =0 → �x =0 → T x

= S(0, 0) + AdT → + (b cx→)dx→ 0 0 − � c� = S(0, 0) + AT + bx x 2 . − 2

However, S(T = 0, x) = S(0, 0) + bx cx2/2, now explicitly depends on x, in violation of − the third law of thermodynamics. (d) Calculate the heat capacity at constant tension, i.e. C = T γS/γT , as a function J |J of T and J. Writing the entropy as S(T, x) = S(T, x(T, J)), leads to •

γS γS γS γx = + . γT γT γx γT �J �x �T �J � � � � � � � � � � 18� � �

From parts (a) and (b), εS = b cx and εS x = A. Furthermore, εx is given by εx T − εT εT J a εx b + cx + cT εx = 0, i.e. εT εT � � � − � γx b cx� � = − . γT a + cT Thus (b cx)2 CJ = TA + − . (a + cT ) � � J +bT Since x = a+cT , we can rewrite the heat capacity as a function of T and J , as

J +bT 2 (b c a+cT ) CJ = TA + − � (a + cT ) �

(ab cJ )2 = TA + − . (a + cT )3 � � ******** 2. Random matrices: As a model for energy levels of complex nuclei, Wigner considered N N symmetric matrices whose elements are random. Let us assume that each element × M (for i j) is an independent random variable taken from the probability density ij � function 1 p(Mij) = for a < Mij < a , and p(Mij) = 0 otherwise. 2a −

(a) Calculate the characteristic function for each element Mij . Since each element is uniformly distributed in the interval ( a, a), the characteristic • − function is a ika ika ikx 1 e e− sin ak p˜ij (k) = dx e− = − = . a 2a 2aik ak �−

(b) Calculate the characteristic function for the trace of the matrix, T tr M = M . ∞ i ii The trace of the matrix is the sum of the N diagonal elements which are independent • � random variables. The characteristic function for the sum of independent variables is simply the product of the corresponding characteristic functions, and thus

N N sin ak p˜T (k) = p˜ii(k) = . ak i=1 � � �

19 Since the trace is the sum of N 1 independent random variables, its cumulants are • √ simply N times those of a single element. The leading cumulants are

T = N M = 0, � ≡c � ij≡ and a x2 a 2 T 2 = NM 2 = N dx = N. c ij a 2a 3 �− � � � � For the qunatity t = T/⇒N, higher order cumulants vanish in the limit of N , and � � thus tr M 3t2 3 lim p t = = exp . N ⇒N −2a2 2αa2 ��

(d) For large N, each eigenvalue � (� = 1, 2, , N) of the matrix M is distributed � ··· according to a probability density function

2 �2 p(�) = 1 2 for �0 < � < �0, and p(�) = 0 otherwise, α�0 � − �0 −

(known as the Wigner semi-circle rule). Find the variance of �.

(Hint: Changing variables to � = �0 sin χ simpliﬁes the integrals.) The mean value of � is zero by symmetry, and hence its variance is given by •

�0 2 �2 �2 = d� �2 1 . c 2 �0 α�0 � − �0 �− � � In the integral, change variables to � = �0 sin χ and d� = �0 cos χdχ, to get

2�2 λ/2 �2 λ/2 �2 λ/2 �2 �2 = 0 dχ cos2 χ sin2 χ = 0 dχ sin2 2χ = 0 dχ (1 cos 4χ) = 0 . c α λ/2 2α λ/2 4α λ/2 − 4 �− �− �− � �

2 2 (e) If in the previous result, we have �0 = 4Na /3, can the eigenvalues be independent of each other? The trace of a matrix is related to its eigenvalues by • N N T = � , = T 2 = �2 + � � . � ≥ c � � � � ≡ �=1 �=1 �=� � � � � � � �√ The cross-correlations of eigenvalues thus satisfy

N 2 2 2 2 a 4Na �� = T � = N N = 0. � ≡ c − � 3 − × 3 � �=� �=1 �√ � � � � � 20 Clearly, this is inconsistent with independent eigenvalues. In fact, the well known result that eigenvalues do not cross implies a repulsion between eigenvalues which leads to a much wider distribution than would result from independent eigenvalues. ******** 3. Viscosity: Consider a classical gas between two plates separated by a distance w. One plate at y = 0 is stationary, while the other at y = w moves with a constant velocity

vx = u. A zeroth order approximation to the one particle density is,

0 n 1 2 2 2 f (p,τ τ q ) = exp (px m�y) + p + p , 1 3/2 −2mk T − y z (2αmkB T ) � B � � � obtained from the uniform Maxwell–Boltzmann distribution by substituting the average value of the gas velocity at each point. (� = u/w is the velocity gradient, while n and T are constants.) (a) The above approximation does not satisfy the Boltzmann equation as the collision term 0 (right hand side of the equation) vanishes, while (the left hand side) df 1 /dt = 0. Find 1 � a better approximation, f1 (pτ ), by considering the linearized Boltzmann equation in the single collision time approximation, i.e. 1 0 1 γ pτ γ 0 f1 f1 f 1 + f 1 − , L ∝ γt m · γτq ∝ − ε � � × � ⎬ where ε is a characteristic mean time between collisions. × We have • γ pτ γ 0 � 0 + f = py(px m�y)f , γt m · γτq 1 mk T − 1 � � B whence 1 0 � f1 = f1 1 ε py (px m�y) . − × mk T − � B ⎫

(b) Calculate the off-diagonal component Pxy (y) of the pressure tensor. The pressure tensor is P (y) = nm c c = n p p /m. From the first order density, • �� ≡ � � � ≡ the off-diagonal element is calculated as

3 px py 1 Pxy(y) = d p f (y) m 1 � 3 px py ε � 0 = d p × p (p m�y)f m −mk T y x − 1 � � B � 2 2 (p m�y) py exp x − exp ε �n 2 2mkB T 2 2mkB T = − × dp (p m�y) − dp p − 2 x x � � � y y � � ⎨ m kB T � − ⇒2αmkB T ⎨ ⇒2αmkB T �� ⎩⎧ �� ⎩⎧ ε �n 2 × = 2 ⎭(mkB T ) = �nε kB T. �⎭ ⎧ �⎧ −m kB T − ×

21 (c) The gas exerts a transverse force per unit area F = P (y = w) on the moving plate. x − xy Calculate this force, and hence obtain the coeﬃcient of viscosity, deﬁned by σ = Fx /�. The pressure tensor calculated in part (b) is in fact independent of the position y, and • the force exerted on the top plate (or the bottom plate) is thus

Fx = Pxy = �nε kB T. − × The coeﬃcient of viscosity is then simply

Fx σ = = nε kB T. � ×

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